Question

a genetic experiment involving peas yielded one sample of offspring consisting of 438 green peas and 120 yellow peas. use a 0.01 significance level to test the claim that under the same circumstances, 24% of offspring peas will be yellow. identify the null hypothesis, alternative hypothesis, test statistic, p - value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. use the p - value method and the normal distribution as an approximation to the binomial distribution. what are the null and alternative hypotheses? a. $h_0:p = 0.24$ $h_1:plt0.24$ b. $h_0:p = 0.24$ $h_1:pgt0.24$ c. $h_0:p
eq0.24$ $h_1:pgt0.24$ d. $h_0:p = 0.24$ $h_1:p
eq0.24$ e. $h_0:p
eq0.24$ $h_1:p = 0.24$ f. $h_0:p
eq0.24$ $h_1:plt0.24$ what is the test statistic? $z=square$ (round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion

The total number of peas \(n = 438+120=558\). The number of yellow peas \(x = 120\). The sample proportion \(\hat{p}=\frac{x}{n}=\frac{120}{558}\approx0.215\)

Step2: Calculate the test statistic

The formula for the test statistic \(z\) in a proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\), where \(p = 0.24\) (the proportion in the null hypothesis), \(\hat{p}=0.215\), and \(n = 558\)

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Answer:

\(z\approx - 1.38\)